What is the Point, and Meaning, of the Mean Value of a Function? Ok, so I know that the mean value of a function, $f(x)$, on the interval $[a,b]$ is given by (or defined by?)
$$\frac{1}{b-a}\int_a^bf(x)~dx$$
but I have $2$ basic questions about this:
$1$: From a purely mathematical point of view, does this have any practical use? Does it give us any extra weapon to add to our mathematical arsenal?
$2$: How excatly can I interpret this as a mean value? I've seen the geometrical interpretation (looking at the area under the graph within the given interval), but I still don't understand how it links to a mean value. For a finite set of values I divide the sum of the values by however many values I have to obtain the mean, but in this case there are infinitely many values, so how is this a mean value?
Thanks for your help.
 A: Let $f:[a,b] \to \Bbb R$ be integrable. Consider the equidistant partition of $[a,b]$ into $n$ subintervals: $$\mathcal{P}_n: \quad a < a + \frac{b-a}{n} < a + 2\frac{(b-a)}{n} < \cdots < a + (n-1)\frac{(b-a)}{n} < b.$$The length of every subinterval is $(b-a)/n$. Then by definition of integral, we have $$\lim_{n \to +\infty} \sum_{k=1}^n f\left(a+k\frac{(b-a)}{n}\right) \frac{b-a}{n} = \int_a^b f(x)\,{\rm d}x,$$ and hence $$\lim_{n\to +\infty} \frac{1}{n} \sum_{k=1}^n f\left(a+k\frac{(b-a)}{n}\right) = \frac{1}{b-a} \int_a^bf(x)\,{\rm d}x.$$The thing inside the limit is the arithmethic mean of the values of $f$ on right-endpoints of the intervals in the partition. And the integral is a limit of means, i.e., a "continuous" mean of $f$ on the entire interval $[a,b]$. Of course, you can do the same with left-endpoints.
Such means are very frequent in Measure Theory, where one replaces the Riemann integral by a Lebesgue integral, and so on. For instance, Lebesgue's differentiation theorem gives the most general statement of what happens when, say, you have intervals of the form $[x_0,x_0+h]$ and want to see what happens with $$\lim_{h \to 0} \frac{1}{h} \int_{x_0}^{x_0+h} f(x)\,{\rm d}x.$$The above is the mean of $f$ on the interval $[x_0,x_0+h]$ and, under suitable assumptions, this limit equals $f(x_0)$, as one might expect.
One tool for proving theorems of this type is the so-called Hardy-Littlewood maximal operator, which is defined in terms of means. And so on and so on.
Bottom line: keep studying analysis and you will see this appear everywhere.
A: I'd like to complement the answers from @Lee Mosher and @Ivo Terek  with a data compression angle. That's the kind of angle that helped me develop an intuition behind the formal concept.
Your function $f$ typically assumes several values on $[a,b]$, but if you wanted to summarize it with ${\bf one}$ number $v$, then what number would you pick?
There are several possibilities, and surely some numbers are better than others (for instance,  you wouldn't pick a negative number if your function is known to assume positive values). Intuitively, you'd want a value that's right in the middle of the values assumed by $f$. Among all possibilities, the mean value is a really good choice. The reason is that, among all the other possible values you could have picked, the mean minimizes the (quadratic) distance with the values taken by the function:
$$v = \arg_u\min \int_a^b |f(x)-u|^2dx$$
In other words, it is the value that is closest to all values of $f$ on that interval.
So, if you view your function as a signal, then an obvious way to compress (summarize) it would be to take the mean value. There are other interpretations for the mean value. In physics, this mean value represents the center of mass of a solid. In signal processing, it is the first mode of vibration of a complex wave (first Fourier mode). In statistics and probabilities, if you keep drawing a random variable (e.g. rolling a dice) and you look at the average of all the numbers you get, it will converge towards the mean of the distribution (law of large numbers). In data mining, the mean value of a data set is its centroid. And the list goes on and on...
A: You wrote:

For a finite set of values I divide the sum of the values by however many values I have to obtain the mean...

Let's look at a Riemann sum for the expression $\frac{1}{b-a} \int_a^b f(x) \, dx$, namely:
$$\frac{1}{b-a} \sum_{i=1}^n f(x_i) \, \Delta x = \frac{1}{b-a} \sum_{i=1}^n f(x_i) \, \frac{b-a}{n} = \frac{\sum_{i=1}^n f(x_i)}{n}
$$
Surprise! It's a finite sum of values divided by the number of values.
Now the amazing thing is that if you study what happens to the numerical values of these finite mean expressions as $n \to \infty$, the absolutely amazing thing is that it approaches a limit (assuming $f : [a,b] \to \mathbb R$ is continuous). That limit has its own special notation
$$\frac{1}{b-a} \left(\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \, \Delta x \right) = \frac{1}{b-a} \int_a^b f(x) \, dx
$$
And for this reason, that last expression is called the mean value of $f$ itself, rather than just the mean value of a selected finite number of values of $f$.
